Write the first three terms of the following geometric sequence: tn = 3(1/2)n-1.
n Tn
1 3(1/2)^0 = 3
2 3(1/2)^1 = 3(1/2) = 3/2
3 3(1/2)^2 = 3(1/4) = 3/4
what's the problem? They gave you the formula; just plug in values for n.
To find the first three terms of the geometric sequence tn = 3(1/2)n-1, we can substitute values of n into the formula.
For the first term (n = 1):
t1 = 3(1/2)^(1-1)
= 3(1/2)^0
= 3(1)
= 3
So, the first term of the sequence is 3.
For the second term (n = 2):
t2 = 3(1/2)^(2-1)
= 3(1/2)^1
= 3(1/2)
= 3/2
So, the second term of the sequence is 3/2.
For the third term (n = 3):
t3 = 3(1/2)^(3-1)
= 3(1/2)^2
= 3(1/4)
= 3/4
So, the third term of the sequence is 3/4.
Therefore, the first three terms of the given geometric sequence are 3, 3/2, and 3/4.
To find the first three terms of the geometric sequence, we need to substitute the values of n in the given formula.
The formula for a geometric sequence is generally written as: tn = ar^(n-1), where tₙ is the nth term, a is the first term, and r is the common ratio.
In this case, the formula is given as tn = 3(1/2)^(n-1).
Let's find the values of t₁, t₂, and t₃ by substituting the corresponding values of n:
t₁ = 3(1/2)^(1-1) = 3(1/2)^0 = 3(1) = 3
t₂ = 3(1/2)^(2-1) = 3(1/2)^1 = 3(1/2) = 3/2
t₃ = 3(1/2)^(3-1) = 3(1/2)^2 = 3(1/4) = 3/4
Therefore, the first three terms of the geometric sequence are 3, 3/2, and 3/4.